Stars

A star is a ball of gas, so why doesn’t it collapse under its own weight. Yup! The outward pressure of the hot stuff pushing outwards! But how much hot stuff does it take to reach a Hydrostatic Equilibrium?

A stars density and pressure increases towards its centre and for it to be stable, that outward pressure must balance the inward pull of gravity.

For a sphere,

∫0R 4πr3(dP/dr) = [P(r) 4πr3]0R – 3∫0R P(r)4πr2 dr

Complicated looking I know, but pretty much summed up by saying that the average pressure is,

<P> = -U/3V

i.e. The average pressure is one third of the gravitational potential energy density.

In most stars, the dominant particles are much slower than c (the speed of light) and are therefore non-relativistic. Remembering (as you do) that the velocity of a particle is v = (vx,vy,vz) and momentum is p = (px,py,pz); for a box with sides of length L, that contains N particles,

px = (N/L3)<pxvx>

And if the particle motions are isotropic (they very likely are as they are random),

<pxvx> = <p∙v> /3

So for a non-relativistic gas, p∙v =mv2 , ρ =-U/3v and Etot = -U/2

However for a relativistic gas, p∙v =pc , ρ =-U and Etot = 0

∴ A relativistic gas (as would be found in a white dwarf or very luminous star) is gravitationally barely bound and unstable. The addition of even a small amount of energy would make the star gravitationally unbound. The Free Fall Collapse Time would be

tff = √(3π/32Gρ) (a function of ρ only!)

Stellar Origin

For a cloud with mass M, radius R, total number of particles N, average particle mass m̅, and temperature T; the gravitational potential is,

U = -C(GMNm̅/R)

The constant C depends on the gas distribution within the cloud, but will be close to 1 for most purposes. The total kinetic energy is,